Physics POTENTIAL ENERGY OF A POINT CHARGE

Potential Energy of A Point Charge

`text(Electric Potential and Potential Energy :)`

Electric potential (represented by symbol V) due to a point charge or a charge configuration at a point is defined as

`V=U/q=` P. E. per unit test charge.

`U=Vq`

In other words it is amount of work done by external agency to bring a unit positive charge from reference point (usually taken at infinity) to given point.

`text(Electric Potential due a Point Charge :)`

In the electric field of a point charge `Q` electric potential at a pont will be

`V(vecr)=((kQq)/r)/q=(kQ)/r`

Potential is a scalar quantity.Electric potential due to a positive charge is taken to be positive and that due to a negative charge is taken to be negative. The potential at a point due to more than one charge can be found simply by adding the potentials due each charge separately.

Plot of electric potential (See Figure)

Electric Potential and Energy due to a Charged Ring

A charge Q is unifonnly distributed over the circumference of a ting. Let us calculate the electric potential at an axial point at a distance r from the centre of the ring.
The electric potential at P due to the charge element dq of the ting is given by

`dV=1/(4piepsilon_0)(dq)/Z=1/(4piepsilon_0)(dq)/(R^2+r^2)^(1/2)`

Hence, the electric potential at P due to the uniformly charged ting is given by

`V=int1/(4piepsilon_0)(dq)/(R^2+r^2)^(1/2)=1/(4piepsilon_0)1/(R^2+r^2)^(1/2)intdq`

`V=1/(4piepsilon_0)Q/sqrt(R^2+r^2)`

And we know `U=Vq`

Electric Potential and Potential Energy Due to a Charged Disc

A non-conducting disc of radius 'R' has a unifonn surface charge density `sigma C//m^2` Let us calculate the potential at a point on the axis of the disc at a distance ' r ' from its centre. The symmetry of the disc tells us that the appropriate choice of element is a ring of radius x and thickness dx. All points on this ring are at the same distance `Z=sqrt(x^2+r^2)`, from the point P. The charge on the ring is `dq = sigmadA = sigma(2pix dx)` and so the potential due to the ring is

`dV=1/(4piepsilon_0)(dq)/Z=1/(4piepsilon_0)(sigma(2pix dx))/sqrt(x^2+r^2)`

Since potential is scalar
The potential due to the whole disc is given by

`V=sigma/(2epsilon_0)int_0^R(xdx)/sqrt(x^2+r^2)`

`V=sigma/(2epsilon_0)[(R^2+r^2)^(1/2)-1]`

Let us see this expression at large distance when r >> R.

`V=1/(4piepsilon_0)Q/r` where `Q = pir^2sigma` is the total charge on the disc.

Thus, we conclude that at large distance, the potential due to the disc is the same as that of a point charge Q.